so say you have 8 things

A B C D E F G H

And then you are going to pick 3 at random

if the one you want is "D", what is the probability of you getting "D"?

11/19/2005 10:30:29 PM

wait fo it

[Edited on November 19, 2005 at 10:44 PM. Reason : pls hold]

11/19/2005 10:32:35 PM

No replacement.

But that isn't right. Fuck, 9th grade math was a long long time ago.

11/19/2005 10:37:21 PM

why isn't it just 3/8??

you are picking D..which is 1 out of 8...
and you are drawing 3 items.......so 3/8???

11/19/2005 10:40:19 PM

i got 3/8

7C2 / 8C3

nCx = n!/x!*(n-x)!

11/19/2005 10:41:38 PM

The total number of outcomes is 56, right?

So how many of those don't have "D"?

God I feel stupid.

11/19/2005 10:41:41 PM

you can only end up picking 8 different items.... you are putting your hand in a bag of 8 items and drawing out 3 items hoping you will get a D...
so it is 3/8
that's the way i'm seeing it

[Edited on November 19, 2005 at 10:43 PM. Reason : yea]

11/19/2005 10:42:26 PM

i agree ^

11/19/2005 10:43:21 PM

No, the total number of outcomes is 56 (8C3).

ABC
ABD
ABE
ABF
ABG
ABH

ACD
ACE
ACF
.....

etcetera

11/19/2005 10:44:16 PM

i must be reading the question wrong then...
i thought he meant he had a bag of 8 letters... he wanted to reach into the bag and draw out 3 letters at the same time
what is the probability he would come out with a D among those 3 letters...
that is 3/8 i think

11/19/2005 10:45:04 PM

so 37.5% chance that a peice of the prophecy set gear will drop on each boss in MC?

11/19/2005 10:45:14 PM

1/8 + 1/7 + 1/6

You have a 1 in however many there are chance each time.

You add because it's an "or" situation. You can pick D on the first or the second or the third try, it doesn't matter. You do not multiply cause you don't have to pick D all three times.

11/19/2005 10:46:05 PM

but u have a 1/8 chance of getting D the first time


say you pull C.


now there's 7 things left in the bag.


so ur chances are now 1/7.


this time you pull F.


so you have C, F, and a third pull.


at this point you have a 1/6 chance of getting D


so you can either pull D or not get it


this time around you get A.


so theres 8 possible outcomes, 7 possible outcomes, then 6 possible outcomes.


ahh okay that makes sense ^

1/8

1/7

1/6

7/24 + 1/7

49/168 + 24/168

63/168

21/52

[Edited on November 19, 2005 at 10:50 PM. Reason : is that it?]

11/19/2005 10:48:21 PM

^^ Incorrect. That would mean that if he picked all eight there would be a (1 / 8) + (1 / 7) + (1 / 6) + (1 / 5) + (1 / 4) + (1 / 3) + (1 / 2) + (1 / 1) = 2.71785714 probability of drawing some particular letter.

11/19/2005 10:49:50 PM

don't hate 'em dude..you'll be facing problems like these every minute when you're making millions in your own buisness.

11/19/2005 10:50:35 PM

man i can't believe i'm thinking so hard about this at 11PM on a saturday night...
i aced ST380 2 years ago, and we had these types of problems...maybe i need to break out the notes aha.

11/19/2005 10:56:23 PM

Hum...maybe.

Isn't the entire number of outcomes far higher than 56? Isn't it 8*7*6*5*4*3*2*1?

11/19/2005 10:57:47 PM

1/8

1/7

1/6

7/24 + 1/7


when you state the 1/7 prob for the second time, that is only in the case that you didn't pull it the first time. Multiply the prob of getting it, times the prob that you're going to be dealing with the delema of having to get it again 1/7*(1-1/8) and add those up sucessivley, then i think it'll be right. but then it should give 3/8 i think.

11/19/2005 10:58:04 PM

1X7X6 / 8X7X6

whats wrong with this one ?

11/19/2005 10:58:28 PM

No, it's 8C8 = 8!/3!5!

8! is the number of ways the set can be ordered.

11/19/2005 10:58:28 PM

Quote :

"whats wrong with this one ?"

maybe b/c you're making up shit...

11/19/2005 11:01:10 PM

^^^That simplifies to 1/8... that is if you only drew 1 time

[Edited on November 19, 2005 at 11:01 PM. Reason : ya]

11/19/2005 11:01:21 PM

it would just be 8 * 7 * 6, because the event (selection of a letter) only occurs three times

which is 336

i guess what's getting me, is the issue of redundancy - does ABC count the same as ACB, as the same as CBA, etc... because in that case here's every way possible to choose D:

DAB - DAC - DAE - DAF - DAG - DAH
DBA - DBC - DBE - DBF - DBG - DBH
DCA - DCB - DCE - DCF - DCG - DCH
DEA - DEB - DEC - DEF - DEG - DEH
DFA - DFB - DFC - DFE - DFG - DFH
DGA - DGB - DGC - DGE - DGF - DGH
DHA - DHB - DHC - DHE - DHF - DHG

thats 42 diff ways to get D.

take out D entirely, I'll do it for A:

ABC - ABE - ABF - ABG - ABH
ACB - ACE - ACF - ACG - ACH
AEB - AEC - AEF - AEG - AEH
AFB - AFC - AFE - AFG - AFH
AGB - AGC - AGE - AGF - AGH
AHB - AHC - AHE - AHF - AHG

that's 30 outcomes, which can be multiplied by 6 (for B, C, E, F, G, H) then added to the previously acquired 30.

Gives 42 options for D, and 210 options for H.

= 42/ (42+210)
= 42/252
=14/84
= 1/6
?

[Edited on November 19, 2005 at 11:07 PM. Reason : wrongs]

rah fuck. the more i try to spatialize this shit the less sense it makes.

[Edited on November 19, 2005 at 11:07 PM. Reason : fdjkldsfdsfklsdfsjklfoi;ewjfoiejfidf BANG BANG BANG]

11/19/2005 11:04:55 PM

is it an african or european D?

11/19/2005 11:05:07 PM

it doesn't matter when you choose the D, as long as it is within those 3 draws....
just the probability of getting a D when you draw 3 letters...
so you could draw D first, then draw 2 more letters..you still drew 3 letters and got a D

11/19/2005 11:06:40 PM

Alright...

If you're taking 3 objects out of eight, and order doesn't matter...

8!7!6
3!2!

...equals 56 possible outcomes. If you were to miss getting D when drawing, you'd have to choose 1 of the other 7 letters first, then 1 of the remaining 6 non-D letters, then finally 1 of the remaining 5 non-D letters. So the probability of NOT choosing D is

7!6!5!
3!2!

...which is 42 35 outcomes. Hence you have a 35/56, or 5/8 chance of NOT choosing D.

Therefore you will pick D 3/8 of the time.

[Edited on November 19, 2005 at 11:22 PM. Reason : I can't multiply...]

11/19/2005 11:09:30 PM

from what iread in your post "3 at a time" would mean ABC == ACB

one at a time and 3 draws like that will set the above two apart

11/19/2005 11:10:00 PM

I like that answer best.

p.s. ORDER DOESN'T MATTER.

11/19/2005 11:10:34 PM

solving of the problem:

thinking like you're pulling 3 at one time:

there's one D. There's 8 things that could be that D. 3 of them are in your hand.

prob := 3 / 8


thinking like you're pulling out 3 sucessivley:

prob getting it first time = 1/8 = .125

prob of not getting it = 1 - 1/8
there are 7 remaining and you have a 1/7 chance of getting it
prob you don't get it and draw it second time = (1-1/8)*1/7 = .125

prob you didn't get it first or second time = 1-1/8-(1-1/8)*1/7
prob you're not gona get it 2 times and get it on the 3rd try = (1-1/8-(1-1/8)*1/7)*1/6 = .125

add all 3 together = .375


doing some stupid shit about possibilities:

# of distint assortments you can get in your hand
8 combination 3
56

# of distint assortments that have a D,
DXX, being that XX is a combination calculation for 2 out of other 7
7 combination 2
21

21/56 = .375


if you get something different. you did it wrong.

[Edited on November 19, 2005 at 11:17 PM. Reason : ]

11/19/2005 11:16:28 PM

Yeah you got it, my logic is faulty somewhere, but I'm not in the mood to figure out why.


Lets say you successfully chose D. You have 7 remaining letters in some two letter combination:

AB AC AE AF AG AH
BC BE BF BG BH
CE CF CG CH
EF EH
FH

Which is 21 combinations. So 21 times the combination will have D. 21/56 = 3/8, which is correct.


Sheesh, I shouldn't have screwed that up.

11/19/2005 11:20:56 PM

3/8 == .375

11/19/2005 11:21:23 PM

so i was right afterall? allllright then...
surprise myself sometimes..
unless someone comes by and proves it wrong again, which is entirely possible..i hate probabilities

[Edited on November 19, 2005 at 11:22 PM. Reason : ya]

11/19/2005 11:21:51 PM

now to problem two

11/19/2005 11:22:27 PM

everyone tired to overthink this, they ignored that order did not matter. all that mattered was picking 3 things and having 1 be d. abd=dab=bad, ect

11/20/2005 1:38:02 AM

this is a college board, right?

11/20/2005 1:43:27 AM

A B C D E F G H

odds you choose D?

(1/8) (no argument here)

now lets pretend you grab two letters; odds to win

(2/8)

grabbing 3;

(3/8)

grabbing 4;

(4/8) (of course its half)


...


grabbing 8;

(8/8) (of course its 100 percent)


[Edited on November 20, 2005 at 2:26 AM. Reason : - ]

11/20/2005 2:17:32 AM

Ok now say you are going to do this 8 times

What is the odds of at least getting D once

and what are the average amount of times you are going to get D?

11/20/2005 3:14:40 AM

1/8 + 1/7 + 1/6

easy

[Edited on November 20, 2005 at 3:34 AM. Reason : ]

11/20/2005 3:34:10 AM

^^ yea if you grab 8 letters then its 100%

^no, thats wrong.

Quote :

"1/8 + 1/7 + 1/6"

lets traslate.

Odds of finding D is 1/8 when you choose your first. But you didnt find D. Therefore you must now choose again. Odds of winning is now 1/7....ok you lose again, now lets try for a third time, its 1/6. the sum of these numbers is now indicates the odds that this scenario plays out.

BUT your numbers assume a case where you do not win the game until the final round. which doesnt happen each time, which is why your way is wrong. you need to subract the probabilities of the cases where you win in round 1 or round 2. do the math like the guy above did. you will find its also 3/8



[Edited on November 20, 2005 at 9:40 AM. Reason : plain english]

11/20/2005 9:28:12 AM

COMBINATIONS AND PERMUATIONS PLEASE

11/20/2005 12:10:02 PM

Quote :

"Ok now say you are going to do this 8 times What is the odds of at least getting D once and what are the average amount of times you are going to get D?"

I'm a little unclear on what you mean. I assume you mean pick one, replace it, pick one, replace it. otherwise it would be a trival question

I belive this is a binomial distribution. p = 0.125, n = 8
expected value for this is E(x) = np = 1

odds of getting D zero are
(n comb x) p^x q^(n-x) = 0.344

where x is 0 and q is 1-p

subtract that number for 1 to get the prob that you did not get zero = the prob you got >= 1, .656

11/20/2005 12:27:32 PM

The odds of you getting D at least once out of eight times is

1 - (3 / 8)^8 = 0.999608934

11/20/2005 12:40:44 PM

questions like this were not meant to take up an entire page

2 posts should have been more than sufficient

11/20/2005 12:57:12 PM

Quote :

"Ok now say you are going to do this 8 times"

Quote :

"The odds of you getting D at least once out of eight times is 1 - (3 / 8)^8 = 0.999608934"

If by "this" it was ment drawing 3 out of the 8 and replacing them for the next time. then by all means yes. but that experiment sucks, there's a 4 in 10000 chance you won't get a D.

just wondering, where did these problems come from?

11/20/2005 2:37:05 PM

Quote :

"Cynic Veteran 497 Posts user info edit post so 37.5% chance that a peice of the prophecy set gear will drop on each boss in MC? 11/19/2005 10:45:14 PM"

11/20/2005 2:50:12 PM

wow..... there is a full page for this common sense question.

11/20/2005 11:44:14 PM

to the purpose of the thread...

Quote :

"so 37.5% chance that a peice of the prophecy set gear will drop on each boss in MC? "

no, it's actually lower than that. even if all the bosses drop 3 epics, that doesn't mean the boss is going to drop 3 set pieces. Further, you can get two of the same set piece from the same boss on occasion. So the algorithm seems to be more similar to selecting 3 pieces from the total loot table, but with replacement. Say most bosses drop two set pieces and then one other epic from the larger loot table, a weapon or ring or random epic armor, etc. So the chances of getting at least one prophecy piece would in theory be more like (1 - 49/64) ~ 23.4%.

11/21/2005 2:05:59 AM

1/8 + 1/7 + 1/6

11/22/2005 3:11:29 PM

the answer is:

you can't pick up the girl sitting in seat D. you can't afford to buy her a drink, she looks like she's already pretty into the guy sitting in C, and it's obvious you're just some liberal arts major with a fake ID anyway. better leave this one to the pros.

11/22/2005 3:17:45 PM